In 140 ml of a 15% sodium nitrite solution (p = 1.04 g / ml), 20 g of sodium

In 140 ml of a 15% sodium nitrite solution (p = 1.04 g / ml), 20 g of sodium nitrate was dissolved. Calculate the mass fraction of salt in the resulting solution.

Given:
V1 solution (NaNO3) = 140 ml
ω1 (NaNO3) = 15%
ρ1 solution (NaNO3) = 1.04 g / ml
m2 (NaNO3) = 20 g

To find:
ω3 (NaNO3) -?

Decision:
1) m1 solution (NaNO3) = ρ1 solution (NaNO3) * V1 solution (NaNO3) = 1.04 * 140 = 145.6 g;
2) m1 (NaNO3) = ω1 (NaNO3) * m1 solution (NaNO3) / 100% = 15% * 145.6 / 100% = 21.84 g;
3) m3 (NaNO3) = m1 (NaNO3) + m2 (NaNO3) = 21.84 + 20 = 41.84 g;
4) m3 solution (NaNO3) = m1 solution (NaNO3) + m2 (NaNO3) = 145.6 + 20 = 165.6 g;
5) ω3 (NaNO3) = m3 (NaNO3) * 100% / m3 solution (NaNO3) = 41.84 * 100% / 165.6 = 25.27%.

Answer: The mass fraction of NaNO3 is 25.27%.